5C (Differential Equations)

W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}

W^{'} = - p W .

W \neq = 0

, the two column vectors in the displayed representation form a basis of

R^{2}

at each

x

. Differentiating the first component and comparing it with the second gives

u^{'} y_{1} + v^{'} y_{2} = 0.

Substitution in the differential equation gives

u^{'} y_{1}^{'} + v^{'} y_{2}^{'} = f .

Solving this two-by-two system yields

u^{'} = - \frac{y _{2} f}{W}, v^{'} = \frac{y _{1} f}{W} .

b

After division by

x^{2}

W = C / x

y_{1} = x^{- 1/2} e^{- x}

allows the convenient choices

y_{2} = x^{- 1/2} e^{x}, W = \frac{2}{x} .

The normalized forcing is

f = x^{3/2}

, so

u^{'} = - \frac{1}{2} x^{2} e^{x}, v^{'} = \frac{1}{2} x^{2} e^{- x} .

Taking convenient antiderivatives and simplifying gives the particular solution

- x^{3/2} - 2 x^{- 1/2}

. Hence

y = C_{1} x^{- 1/2} e^{- x} + C_{2} x^{- 1/2} e^{x} - x^{3/2} - 2 x^{- 1/2} .

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